5(x+3)-5x^2-1=x^2+7(3-x)-1

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Solution for 5(x+3)-5x^2-1=x^2+7(3-x)-1 equation:



5(x+3)-5x^2-1=x^2+7(3-x)-1
We move all terms to the left:
5(x+3)-5x^2-1-(x^2+7(3-x)-1)=0
We add all the numbers together, and all the variables
-5x^2+5(x+3)-(x^2+7(-1x+3)-1)-1=0
We multiply parentheses
-5x^2+5x-(x^2+7(-1x+3)-1)+15-1=0
We calculate terms in parentheses: -(x^2+7(-1x+3)-1), so:
x^2+7(-1x+3)-1
We multiply parentheses
x^2-7x+21-1
We add all the numbers together, and all the variables
x^2-7x+20
Back to the equation:
-(x^2-7x+20)
We add all the numbers together, and all the variables
-5x^2+5x-(x^2-7x+20)+14=0
We get rid of parentheses
-5x^2-x^2+5x+7x-20+14=0
We add all the numbers together, and all the variables
-6x^2+12x-6=0
a = -6; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·(-6)·(-6)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-12}{-12}=1$

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